Rút gọn phân thức a^3-3a+2/2a^3-7a^2+8a-3
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Những câu hỏi liên quan
6 tháng 7 2016
Tử = \(a^3-3a+2=a^3-1-3a+3\)
\(=\left(a-1\right)\left(a^2+a+1\right)-3\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2+a-2\right)\)
\(=\left(a-1\right)\left(a-1\right)\left(a+2\right)=\left(a-1\right)^2\left(a+2\right)\)
Mẫu =\(2a^3-7a^2+8a-3=2a\left(a^2-2a+1\right)-3\left(a^2-2a+1\right)\)
\(=\left(a-1\right)^2\left(2a-3\right)\)
=>\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}=\frac{\left(a-1\right)^2\left(a+2\right)}{\left(a-1\right)^2\left(2a-3\right)}=\frac{a+2}{2a-3}\)
Nhớ h cho mik nhé
LM
19 tháng 7 2017
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
LM
19 tháng 7 2017
b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
TL
1
6 tháng 8 2019
\(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-36a^2+24a-4\)
\(=-7a^2+4a-7\)
27 tháng 11 2016
\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)
\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)
Câu b xem lại đề đi nhé
NV
0
\(\dfrac{a^3-3a+2}{2a^3-7a^2+8a-3}\)
\(=\dfrac{a^3-a-2a+2}{2a^3-2a^2-5a^2+5a+3a-3}\)
\(=\dfrac{a\left(a-1\right)\left(a+1\right)-2\left(a-1\right)}{2a^2\left(a-1\right)-5a\left(a-1\right)+3\left(a-1\right)}\)
\(=\dfrac{\left(a-1\right)\left(a^2+a-2\right)}{\left(a-1\right)\left(2a^2-5a+3\right)}\)
\(=\dfrac{\left(a+2\right)\left(a-1\right)}{\left(a-1\right)\left(2a-3\right)}\)
\(=\dfrac{a+2}{2a-3}\)